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Let's have the following sequence 43, 108, 5737, 13932, 97613377, 239783652,?

What number belongs where the question mark is?

HINT: The next number is 28693771291496497. What is the next term after that?

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I believe the next few numbers are

28693771291496497 (as given in the hint), 70218276063338412, 2466867942031109947559641241498257, 519657642496019930838140921497014645213420297443652.

And, just for fun, the previous numbers are

1, 3, 4, 9.

If we write

the sequence as $a_1,b_1,a_2,b_2,\ldots$ (so that, empirically, we have roughly $b_j=\sqrt{6}a_j$ and $a_{j+1}=\frac{\sqrt6}2b_j$)

then it appears that

$b_{j+1}=3a_jb_j$ exactly, and $a_{j+1}\simeq3a_j^2$. If we write $d_j=a_{j+1}-3a_j^2$ so that $d_1=5737-3\cdot43^2=190$ and $d_2=97613377-3\cdot5737^2=-1126130$, we may notice that $d_2$ is a multiple of $d_1$, and the quotient is rather close to $-a_2$; in fact, for the given numbers we find that $d_{j+1}=-d_j(a_{j+1}+d_j)$.

So, successively, we find (beginning after the originally given sequence and thus confirming the value in the hint):

$d_3=-d_2(a_3+d_2)=-(-1126130)\cdot(97613377+(-1126130))=108657183464110$
$a_4=3a_3^2+d_3=3\cdot97613377^2+108657183464110=28693771291496497$
$b_4=3a_3b_3=3\cdot97613377\cdot239783652=70218276063338412$
$d_4=-d_3(a_4+d_3)=-108657183464110\cdot(28693771291496497+108657183464110)=-3129590755015700672104048314770$
$a_5=3a_4^2+d_4=2466867942031109947559641241498257$
$b_5=3a_4b_4=519657642496019930838140921497014645213420297443652$

and so forth.

As a minor (equivalent) variation,

instead of $3a_j^2$ we could write $a_jb_{j+1}/b_j$ which is the same because $b_{j+1}=3a_jb_j$.

What terms precede the given ones? We have $a_0,b_0,43,108,\ldots$ so

$108=b_1=3a_0b_0$ so $a_0b_0=36$;
$d_0=a_1-3a_0^2$ so $d_0=43-3a_0^2$;
$d_1=-d_0(a_1+d_0)$ so $190=-d_0(43+d_0)$;
the last of these is a quadratic equation in $d_0$ whose two solutions are $d_0=-5$ and $d_0=-38$;
only the first of these makes $a_0$ an integer, namely $\pm4$,
and then $b_0=\pm9$. So presumably we have $4,9,43,108,\ldots$.

Can we go further back?

$9=b_0=3a_{-1}b_{-1}$ so $a_{-1}b_{-1}=3$;
$d_{-1}=a_0-3a_{-1}^2$ so $d_{-1}=4-3a_{-1}^2$;
$d_0=-d_{-1}(a_0+d_{-1})$ so $-5=-d_{-1}(4+d_{-1})$;
the last of these is a quadratic equation in $d_{-1}$ with solutions $d_{-1}=1,-5$;
only the former solution makes $a_{-1}$ an integer, namely 1;
and then $b_{-1}=3$.
So we have $1,3,4,9,43,108,\ldots$.

Further?

$3=b_{-1}=3a_{-2}b_{-2}$ so $a_{-2}b_{-2}=1$;
$d_{-2}=a_{-1}-3a_{-2}^2$ so $d_{-2}=1-3a_{-2}^2$;
$d_{-1}=-d_{-2}(a_{-1}+d_{-2})$ so $1=-d_{-2}(1+d_{-2})$;
the last of these is a quadratic equation in $d_{-1}$ whose solutions are not integers, or indeed real numbers. So we'd probably better stop there.

If there's a nice closed form for the numbers in this sequence, it's not apparent to me. I am curious where (if anywhere) it comes from. (That is: what mathematical question or construction, if any, is associated with it in something like the way that rational approximations to $\frac{1+\sqrt5}2$ are associated with the Fibonacci sequence.)

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  • $\begingroup$ I said previously that all my postings are of my own creation. The number you found for the next term is not correct. Nonetheless, your approach is on the right path. $\endgroup$ – Vassilis Parassidis Mar 30 at 17:47
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    $\begingroup$ Sorry, "where it comes from" didn't mean "what other source you stole the question from"! I meant: this seems like it may be derived from some bit of mathematics, in the same sort of way as simpler recurrence relations can be derived from rational approximation of irrational numbers and the like, and I was wondering what other bit of mathematics was involved. $\endgroup$ – Gareth McCaughan Mar 30 at 18:56
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    $\begingroup$ If I haven't simply miscalculated, and if my claim about how the sequence works is compatible with the numbers you've posted so far, then I think you should either (1) post another term or two or (2) say explicitly (if it's true) that the intended solution is much simpler than mine. (If it isn't, and if my calculations aren't outright wrong, then that indicates that the numbers given aren't sufficient to pick out a specific sequence, which is why you should post another term or two.) $\endgroup$ – Gareth McCaughan Mar 30 at 18:59
  • $\begingroup$ On the other hand, if I just slipped up in my calculations, do let me know and I'll double-check them :-). $\endgroup$ – Gareth McCaughan Mar 30 at 19:00
  • $\begingroup$ I added the next number as a hint. $\endgroup$ – Vassilis Parassidis Mar 30 at 19:52

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